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    hey guys,

    I really can't do this question and need some help!

    \displaystyle \int \frac{1}{(1-x^2)^{3/2}}\ \mathrm{d}x


    limits are x=1/2, x=0

    and thats (1-x^2)^3/2 I'm noob at latex

    So they say that x=sinu.

    therefore (1-x^2)^3/2 = cos^2u

    dx= cosu

    Now *imagine integral sign here* 2/cos2u+1

    Now surely after intergrating that you would get ln (sin2u +1) ?
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    PS Helper
    This is a joke, right?

    Also, LaTex fail...
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    (Original post by J DOT A)
    hey guys,

    I really can't do this question and need some help!


    \int 1/ (1-x)^2 dx

    limits are x=1/2, x=0
    Try writing it as \int (1-x)^{-2} dx and integrating by inspection through noting that \int (ax+b)^n dx = \frac{(ax+b)^{n+1}}{a(n+1)} + c, where a and b are constants.

    If not, use a substitution of u=1-x to start with.

    EDIT due to incorrect LaTeX in the OP...: If the question is giving you the substitution then what more can we help without you at least posting some working?
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    (Original post by J DOT A)
    hey guys,

    I really can't do this question and need some help!


    \int 1/ (1-x)^2 dx

    limits are x=1/2, x=0
    Well consider what you would need to differentiate to get that and adjust any numerical factors, if you need... If you still don't see it, try a substitution if you reallllly have to. But doing it by inspection will save you a lot of time.


    (Original post by J DOT A)
    hey guys,

    I really can't do this question and need some help!


    \int 1/ (1-x^2)^3 divided by 2 dx

    limits are x=1/2, x=0

    And they give the substituion as x=sinu

    Okay so
    Here's an idea you may like to try: Use the preview button.
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    err..
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    (Original post by Doughnuts!!)
    This is a joke, right?

    Also, LaTex fail...
    Sorry guys!!! I am such a noob at latex, this is my first time
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    (Original post by dknt)
    Well consider what you would need to differentiate to get that and adjust any numerical factors, if you need... If you still don't see it, try a substitution if you reallllly have to. But doing it by inspection will save you a lot of time.




    Here's an idea you may like to try: Use the preview button.
    LOL I will do that next time fml:P
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    (Original post by J DOT A)
    hey guys,

    I really can't do this question and need some help!


    \int 1/ (1-x^2)^3 divided by 2 dx

    limits are x=1/2, x=0

    and thats (1-x)^3/2 I'm noob at latex

    So they say that x=sinu.
    LaTeX you want is [latex]\displaystyle \int \frac{1}{(1-x^2)^{3/2}}\ \mathrm{d}x[/latex] : \displaystyle \int \frac{1}{(1-x^2)^{3/2}}\ \mathrm{d}x

    Yeah so sub in x = sinu, work out "dx", sub that in and replace all the x terms with u terms. Remember your basic trig identities.

    Although if you mean \frac{1}{(1-x)^{3/2}} then ignore me.
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    (Original post by Doughnuts!!)
    This is a joke, right?

    Also, LaTex fail...
    A little harsh, don't you think? :p:

    For some reason, a lot of people doing C4 tend to see trig substitution as more difficult than a polynomial substitution. Not sure why, but still.
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    (Original post by Farhan.Hanif93)
    Try writing it as \int (1-x)^{-2} dx and integrating by inspection through noting that \int (ax+b)^n dx = \frac{(ax+b)^{n+1}}{a(n+1)} + c, where a and b are constants.

    If not, use a substitution of u=1-x to start with.

    EDIT due to incorrect LaTeX in the OP...: If the question is giving you the substitution then what more can we help without you at least posting some working?
    Sorry for this piss up with the latex
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    (Original post by J DOT A)
    LOL I will do that next time fml:P
    Hehe, don't worry about it We all gotta start somewhere!
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    (Original post by J DOT A)
    Sorry for this piss up with the latex
    No need to apologise! :p:

    Feel free to post up your working and we'll try to help you become unstuck - it's a bit tough to help you without seeing working, especially if the question gives you the required substitution.
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    (Original post by Swayum)
    LaTeX you want is [latex]\displaystyle \int \frac{1}{(1-x^2)^{3/2}}\ \mathrm{d}x[/latex] : \displaystyle \int \frac{1}{(1-x^2)^{3/2}}\ \mathrm{d}x

    Yeah so sub in x = sinu, work out "dx", sub that in and replace all the x terms with u terms. Remember your basic trig identities.
    Thanks for that

    and yeah I posted my working in the OP
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    (Original post by Farhan.Hanif93)
    No need to apologise! :p:

    Feel free to post up your working and we'll try to help you become unstuck - it's a bit tough to help you without seeing working, especially if the question gives you the required substitution.
    ahh thanks aha

    Yeah just posted my working in the OP
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    (Original post by Farhan.Hanif93)
    A little harsh, don't you think? :p:

    For some reason, a lot of people doing C4 tend to see trig substitution as more difficult than a polynomial substitution. Not sure why, but still.
    Haha, yeah, I came off as a bit mean. And now that they've amended it, I can see why they'd ask for help.

    I'm one of those people! Trig substitution can get a bit messy at times.
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    (Original post by J DOT A)
    hey guys,

    I really can't do this question and need some help!


    \int \dfrac{1}{(1-x^2)^{\frac{3}{2}}}dx

    limits are x=1/2, x=0

    and thats (1-x^2)^3/2 I'm noob at latex

    So they say that x=sinu.

    therefore (1-x^2)^3/2 = cos^2u

    dx= cosudu

    Now *imagine integral sign here* 2/cos2u+1

    Now surely after intergrating that you would get 2ln (sin2u +1) ?
    I've highlighted any notation that you missed out in red and the bolded lines are the ones which you have made mistakes in.

    For the first line, think about what power cosx should have after you've fully simplified everything. Notice that (1-sin ^2x)^{\frac{3}{2}} = (\sqrt{\cos ^2 x})^3.

    For the other bolded lines, they are really completely wrong so I'm not sure where to start with those ones. You should really reread your textbook, in particular the chapters about integration by trigonometric substitution and trigonometric identities.
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    (Original post by Farhan.Hanif93)
    I've highlighted any notation that you missed out in red and the bolded lines are the ones which you have made mistakes in.

    For the first line, think about what power cosx should have after you've fully simplified everything. Notice that (1-sin ^2x)^{\frac{3}{2}} = (\sqrt{\cos ^2 x})^3.

    For the other bolded lines, they are really completely wrong so I'm not sure where to start with those ones. You should really reread your textbook, in particular the chapters about integration by trigonometric substitution and trigonometric identities.
    Ahh i skipped alot out as it was to hard for me to put on latex.

    Basically you get cos^3u from simplifiying (sqrt*cos^2u)^3.
    now this will become cos^2u as you cancel it out with the dx=cosu.
    therefore you get 1/cos^2u

    now using the trig idenitity 1/2(cos2A+1)= cos^2u...

    You get the integral of 2/cos2A+1
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    (Original post by J DOT A)
    Ahh i skipped alot out as it was to hard for me to put on latex.

    Basically you get cos^3u from simplifiying (sqrt*cos^2u)^3.
    now this will become cos^2u as you cancel it out with the dx=cosu.
    therefore you get 1/cos^2u

    now using the trig idenitity 1/2(cos2A+1)= cos^2u...

    You get the integral of 2/cos2A+1
    Ah ok, that makes a lot more sense! :p:

    What I would do is avoid going into using the double angle formulae at all and just recall that \frac{1}{\cos x} \equiv \sec x so you have the integral of \sec ^2x to deal with. If you don't recognise this immediately, what is the derivative of tanx? What can you tell me about the integral of \sec ^2x following that?

    This is one integral that you should really know off by heart.
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    (Original post by Farhan.Hanif93)
    Ah ok, that makes a lot more sense! :p:

    What I would do is avoid going into using the double angle formulae at all and just recall that \frac{1}{\cos x} \equiv \sec x so you have the integral of \sec ^2x to deal with. If you don't recognise this immediately, what is the derivative of tanx? What can you tell me about the integral of \sec ^2x following that?

    This is one integral that you should really know off by heart.
    ahh you absolute LEGEND! Thank you so much for that aha!

    Yeah I just need to recognise stuff more I guess aha

    Once again thank you, and thanks to all that contributed to the thread.
 
 
 
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